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iragen [17]
3 years ago
14

The roots of the equation 3x²-2x-4=0 are J and K. Evaluate J² + K².

Mathematics
1 answer:
melamori03 [73]3 years ago
6 0

Answer:

28/9

Step-by-step explanation:

If the roots are J and K, then:

3 (x − J) (x − K) = 0

3 (x² − (J+K)x + JK) = 0

So if we factor out the leading coefficient:

3x² − 2x − 4 = 0

3(x² − 2/3x − 4/3) = 0

The coefficient of the second term is the sum of the roots:

J + K = 2/3

And the constant is the product of the roots:

JK = -4/3

If we take the sum of the roots and square it:

(J + K)² = (2/3)²

J² + 2JK + K² = 4/9

And subtract twice the product:

J² + K² = 4/9 − 2JK

J² + K² = 4/9 − 2(-4/3)

J² + K² = 4/9 + 8/3

J² + k² = 28/9

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